1000 Solved Problems in Modern Physics

2.3 Solutions 107

mv^2 /r=kr (2)

solving (1) and (2)

v^2 =

(

n

m

)(

k

m

)^12

(3)

r^2 =

n

√

km

(4)

(c) E=U+T=

1

2 kr

(^2) + 1
2 mv
(^2) (5)
Substituting (3) and (4) on (5) and simplifying
E=n(k/m)^1 /^2
(d)ΔE=En−En− 1 =

(

k

m

)^12

= 1. 05 × 10 −^34

(

1 , 180

3 × 10 −^26

)^12

J

= 0 .13 eV

λ=

1 , 241

0. 13

= 9 ,546 nm

2.17 En=−

13. 6

n^2

ΔE=En+ 1 −En= 13. 6

(

1

n^2

−

1

(n+1)^2

)

=

13 .6(2n+1)

n^2 (n+1)^2

In the limitn→∞,ΔE ∝nn 4 =n^13

2.18 The wavelengthλ= 486 .1 nm corresponds to the transition energy ofE=
1241 / 486. 1 = 2 .55 eV Looking up Fig. 2.1, for the energy level diagram
for hydrogen atom, the transitionn = 4 →2 gives the energy difference
− 0. 85 −(− 3 .4)= 2 .55 eV
The line belongs to the Balmer series.

Fig. 2.1Energy level
diagram for hydrogen atom

2.19 Orbital velocity,v= e
2
2 nhε 0 ,a^0 =n

(^2) h (^2) ε 0 /πe (^2) m
Orbital frequencyf=v/ 2 πa 0